3.3 Video 1 Hacks

Show two examples and label which one is sequence, selection, iteration

numbers = [0,1,2,3,4,5,6,7,8,9,10]
evens = []

for i in numbers:
    if (numbers[i] % 2 == 0):
        evens.append(numbers[i])

print(evens)

[0, 2, 4, 6, 8, 10]

The cell as a whole is a sequence.

"for i in numbers" is an iteration.

if (numbers[i] % 2 == 0): is a selection.

3.3 Video 2 Hacks

Practice Problems

  1. given the following code segment below:

a ⟵ 7

b ⟵ 1

c ⟵ 3

d ⟵ 4

a ⟵ b

b ⟵ c + d

d ⟵ b

find the value for a, b, c, d

My answer: a = 1 (a changes from 7 to 1) b = 7 (b changes from 1, to 3+4) c = 3 (c does not change) d = 7 (d changes from 4 to 7, due to b changing)

  1. consider the following code segment:

hot ⟵ true

cold ⟵ false

cold ⟵ hot

hot ⟵ cold

what are the values of hot and cold after executing the code segment?

  1. the value of hot is true, the value of cold is true

My answer:

Cold changes to hot, and hot stays true

  1. Make TWO of your own code segments that contain at least 5 defined variables, then provide the answer and EXPLAIN why your answer is correct.

Segment 1:

red ⟵ yes

orange ⟵ no

green ⟵ almost

blue ⟵ never

orange ⟵ blue

purple ⟵ absolutely

red ⟵ orange

What is red?

never

Segment 2:

a ⟵ 8

b ⟵ 3

c ⟵ 7

d ⟵ 9

a ⟵ 2

b ⟵ a + c

c ⟵ b

Find the value for c.

c = 15

  1. Sequencing
num1 = 3
num2 = 1
num3 = 5
num1 = num2 + num3      
num2 = num1 + num3

What is the value of num1 and num2?

num1 = 6

num2 + num 3 = 6

num2 = 11

num1 + num3 = 6 + 5 = 11

3.4 Video 1 Hacks

String Homework

  • Test 1

    firstName <- "Bob" lastName <- "Smith" var <- substring(firstName, 1, 1) name <- concat(lastName, var) email <- concat(name, "@gmail.com") DISPLAY(email)

  • What would the result be?

substring individualized the letter B, then name is defined as SmithB, the email then adds the address to become:

SmithB@gmail.com

  • Test 2

    word1 <- "computer" word2 <- "textbooks" length1 <- len(word1)/2 length2 <- len(word2)/3 first <- substring(word1, 2, len1) second <- substring(word2, len2+3, len2) newWord <- concat(first, second) DISPLAY(newWord)

ompuook